Explaining bowling ball specs

[IHateTenPins]

I'm still learning about bowling balls and I want to see if I got this right. Low RG compared to High RG is when a ball hooks? So a low Rg ball hooks earlier than a high RG ball?


And differential is Hook potential?

[Mike White]

I'm still learning about bowling balls and I want to see if I got this right. Low RG compared to High RG is when a ball hooks? So a low Rg ball hooks earlier than a high RG ball?


And differential is Hook potential?

Low RG compared to High RG.

Low RG requires less torque to accelerate the angular velocity.

Translated: The ball revs up easier.

That means the lower RG ball STOPS hooking sooner, because when the rev rate matches the ball speed the ball quits hooking.

The ball needs friction to begin hooking, and that's all due to the surface of the ball.

The RG (high or low) has no effect on when the ball begins to hook.

Differential is related to flare potential.

That is connected to how quickly the ball will sense a significant increase in friction as it exits the oil pattern.

A ball with no flare will still be tracking along where there is oil on the surface of the ball when it has exited the oil pattern.

Some of that oil will be re-deposited onto the lane each revolution.

That re-depositing is the cause of carry down.

A ball that flares, but thrown before the oil is absorbed, (or wiped off) will cause more carry down than a non-flaring ball.

[ChuckR]

Doesn't the layout also modify the way the ball will react? a 40x4"x30 has a different result than a 65x5"x50 on the same ball and PAP.

[bowl1820]

RG: Radius Of Gyration.
The RG of a bowling ball tells you how soon the core is designed to roll.
A Low RG ball (aka: a Center Heavy ball) is easier to "rev up", rolls earlier and tends to mean earlier hook, more midlane reaction,a more evenly arcing ball used on wetter conditions.

A High RG ball (aka: a Cover Heavy ball) is harder to "rev up", rolls later and tends to mean later hook, more backend reaction, tends to be more angular.

The Low RG axis is the pin and the High RG axis is the PSA.
Total Diff = High RG - Low RG
Int Diff = High RG - Int RG


The RG Differential aka:The Differential:
Shows the difference between the low RG and the high RG.
The Differential indicates the potential for track flare.
The lower the differential the closer together the track flare rings are.
The higher the differential the further apart the track flare rings are.


Basic guide:
A lower RG ball with a low differential will produce an earlier rolling ball with a small arcing motion.
A lower RG ball with a high differential will produce an earlier rolling ball with a larger/stronger arcing motion.
A higher RG ball with a low differential will produce a later rolling ball with an angular backend motion.
A higher RG ball with a high differential will produce a later rolling ball with a strong, somewhat angular backend motion.

[djp1080]

Low RG compared to High RG.

Low RG requires less torque to accelerate the angular velocity.

Translated: The ball revs up easier.

That means the lower RG ball STOPS hooking sooner, because when the rev rate matches the ball speed the ball quits hooking.

The ball needs friction to begin hooking, and that's all due to the surface of the ball.

The RG (high or low) has no effect on when the ball begins to hook.

Differential is related to flare potential.

That is connected to how quickly the ball will sense a significant increase in friction as it exits the oil pattern.

A ball with no flare will still be tracking along where there is oil on the surface of the ball when it has exited the oil pattern.

Some of that oil will be re-deposited onto the lane each revolution.

That re-depositing is the cause of carry down.

A ball that flares, but thrown before the oil is absorbed, (or wiped off) will cause more carry down than a non-flaring ball.

Mike, I read your response and had to stop at line #4 where you're talking about rev rate matching ball speed.
What caused me to stop, was that rev rate is usually in revs per minute and ball speed is usually in feet per minute; therefore, they can never match.
I'm supposing that there is some magic ratio in values if you're using revs/min and feet/min as your units. What is that magic ratio that you might be referring to???

Thanks...

[Mike White]

Mike, I read your response and had to stop at line #4 where you're talking about rev rate matching ball speed.
What caused me to stop, was that rev rate is usually in revs per minute and ball speed is usually in feet per minute; therefore, they can never match.
I'm supposing that there is some magic ratio in values if you're using revs/min and feet/min as your units. What is that magic ratio that you might be referring to???

Thanks...

1 mile = 5280 feet = 63,360 inches.

The circumfrence of a ball is about 27 inches, (effectively less as you increase axis tilt)

63,360 / 60 (convert to minutes) / 27 (convert to revolutions) = 39.111

A ball rolling 1 mph, will have a rev rate of 39.111 rpm

When you release the ball, it will most likely have more forward speed than equivalent rev rate.

The ball senses friction from the lane causing it to slow down, but since the direction of the friction force is not directly thru the center of the ball, it causes an increase in rev rate.

The best analogy is an airplane tire when it's landing, before contact the wheel isn't spinning as fast as the plane is traveling forward.

When it does make contact, friction causes the wheel to increase it's rev rate until it matches the planes speed.

Take whatever speed the ball is at as it enters the pins, multiply by 39.111, and you have a good estimate of the rev rate as it enters the pins.

This of course assume you threw the ball in a way that it will roll before hitting the pins.

[fordman1]

Mike you should not use about and assuming when talking like the guy in the wheel chair who has an iq of 227.

[Mike White]

Doesn't the layout also modify the way the ball will react? a 40x4"x30 has a different result than a 65x5"x50 on the same ball and PAP.

Layout modifies the resulting RG of the PAP, and modifies the amount of flare possible.

If you put the pin on the PAP, you get no flare, just like if you put the pin 6 3/4" from the PAP.

Drilling angle 40 vs 65 has no effect on a symmetrical ball.

Val Angle can increase or decrease the differential depending on the angle.

Smaller would increase.

The practical limits for val angle are to avoid tracking over the holes.

[Mike White]

Mike you should not use about and assuming when talking like the guy in the wheel chair who has an iq of 227.

I used the word "about" because not all balls have a maximum circumference of 27" due to being resurfaced, and as you increase tilt, the ball will need a higher rpm compared to mph when rolling.

I used the word "assume", because most shots that are thrown completely in the oil are not rolling when contacting the pins, and therefore mph * 39.111 wouldn't give a good estimation of rpm.

[IHateTenPins]

Thanks everyone for their explanations, I think I understand it better now and it'll help me with my next ball purchase.

[fordman1]

Just teasing Mike. All that technical stuff is hard to understand. Remember most of us are once or twice a week league bowler. We go to the pro shop and say give me a strong ball that goes long or what ever our choice is. My eyes glaze over when I read to much tech. You could make up this stuff and most of us wouldn't know the difference.

I have done that with numbers and names. Ernie Kabonze was the 1st bowler who did this or that and no one ever questioned it. Pulled it right out of my A$$.

I was just having a little fun.

[Mike White]

Just teasing Mike. All that technical stuff is hard to understand. Remember most of us are once or twice a week league bowler. We go to the pro shop and say give me a strong ball that goes long or what ever our choice is. My eyes glaze over when I read to much tech. You could make up this stuff and most of us wouldn't know the difference.

I have done that with numbers and names. Ernie Kabonze was the 1st bowler who did this or that and no one ever questioned it. Pulled it right out of my A$$.

I was just having a little fun.

All that technical stuff is fairly basic math.

That reminds me of my neighbor who wanted to get into the local pipe fitters union.

On the entrance test they had some basic math questions. Mostly about converting fractions into decimals, and reverse.

This guy previously was a carpet layer, so using a tape measure wasn't something he had trouble with.

He failed on the first try, so he asked me for some tutoring.

When I looked at his sample test, I said "This test is culturally biased".

It was biased against the culture of students that didn't pay attention in math class.

He passed just fine on the second try, and is now 2 years into a 5 year stint of apprenticeship.

[classygranny]

1 mile = 5280 feet = 63,360 inches.

The circumfrence of a ball is about 27 inches, (effectively less as you increase axis tilt)

63,360 / 60 (convert to minutes) / 27 (convert to revolutions) = 39.111

A ball rolling 1 mph, will have a rev rate of 39.111 rpm

When you release the ball, it will most likely have more forward speed than equivalent rev rate.

The ball senses friction from the lane causing it to slow down, but since the direction of the friction force is not directly thru the center of the ball, it causes an increase in rev rate.

The best analogy is an airplane tire when it's landing, before contact the wheel isn't spinning as fast as the plane is traveling forward.

When it does make contact, friction causes the wheel to increase it's rev rate until it matches the planes speed.

Take whatever speed the ball is at as it enters the pins, multiply by 39.111, and you have a good estimate of the rev rate as it enters the pins.

This of course assume you threw the ball in a way that it will roll before hitting the pins.

So if your ball speed is only 14 mph at the pins, you have approximately a rev rate of 547? Seems like a high rev rate to me, but since I really don't know if this is mathematically correct, I will assume so.

[djp1080]

So if your ball speed is only 14 mph at the pins, you have approximately a rev rate of 547? Seems like a high rev rate to me, but since I really don't know if this is mathematically correct, I will assume so.
Yes, that's the kind of number I get when I put in my estimated ball speed at the pins... :)
Now we need to see if we're actually using the full circumference of the ball. Probably not!
So the magic formula for the casual bowler isn't quite 39.111 now, is it?
Thanks for the pretty good explanation though Mike...

[bowl1820]

So if your ball speed is only 14 mph at the pins, you have approximately a rev rate of 547?
Yes, It's because the rpm's are calculated differently at that end of the lane.

It's why when you look at the ball motion chart it shows max rev's in the roll phase. (Look up the USBC Ball motion study and you can find the math about it.)

http://s5.postimg.org/is5flijd3/Three_Phases_Slide_1.jpg


Seems like a high rev rate to me, but since I really don't know if this is mathematically correct, I will assume so.

MO Pinel had rule of thumb to approx. it, you add about 50% the rpm's you have in the heads (EX. 300 rpms in the heads becomes about 450 rpms at the pins.)

To really calculate it accurately , you need to know where the ball actually starts the roll and get the distance from there to the pins and the time it took to travel that distance, so you can figure out the exact speed. Then calculate the rpms.

[Mike White]

So if your ball speed is only 14 mph at the pins, you have approximately a rev rate of 547? Seems like a high rev rate to me, but since I really don't know if this is mathematically correct, I will assume so.

You're used to associating the ball speed at release with the rev rate at release.

So yes, 14 mph with 547 rpm at release would be a very high rev rate.

The question was when the ball rolls.

If you release the ball at 14 mph, and a rev rate lower than 547, when the ball encounters friction, the speed will decrease, as the rev rate increases until the rev rate (in rpm) is 39.111 times the speed (in mph)

[Mike White]

Yes, that's the kind of number I get when I put in my estimated ball speed at the pins... :)
Now we need to see if we're actually using the full circumference of the ball. Probably not!
So the magic formula for the casual bowler isn't quite 39.111 now, is it?
Thanks for the pretty good explanation though Mike...

The circumference of a circle is 2 * pi * radius.

The radius of the ball at 0 degrees of tilt is 4.2972.

As you increase tilt, the radius of the circle on the surface of the ball that contacts the lane is 4.2972*cos(tilt)

So the full calculation of the MPH to RPM is RPM = MPH * 63360 / (2 * pi * 4.2972 * cos(tilt))

[Mike White]

Yes, It's because the rpm's are calculated differently at that end of the lane.

It's why when you look at the ball motion chart it shows max rev's in the roll phase. (Look up the USBC Ball motion study and you can find the math about it.)

http://s5.postimg.org/is5flijd3/Three_Phases_Slide_1.jpg



MO Pinel had rule of thumb to approx. it, you add about 50% the rpm's you have in the heads (EX. 300 rpms in the heads becomes about 450 rpms at the pins.)

To really calculate it accurately , you need to know where the ball actually starts the roll and get the distance from there to the pins and the time it took to travel that distance, so you can figure out the exact speed. Then calculate the rpms.

There are three pieces of text in that graph that are just plain wrong.

Hook Phase - Force created by the rev rate exceeds the force created by ball speed.

Skid Phase - Force created by ball speed exceeds the force created by the rev rate.

Transition from skid to Hook - Force from speed = force from revs.

In the skid phase, the ball isn't slowing down, and isn't changing direction.

Force = Mass * Acceleration.

If there isn't any acceleration, there is no force.

So force from ball speed = force from rev rate all thru the skid phase, and also all thru the roll phase.

At the beginning of the hook phase, the force from ball speed will exceed force from rev rate.

At the end of the hook phase, both the force from ball speed, and force from rev rate will reach 0.

A ball in the roll phase is not accelerating, and not changing directions so F = M * A says force = 0.


You don't need to know exactly where it rolls, only that it does roll.

Using conservation of momentum, you can calculate the momentum at the release of both the ball speed, and rev rate (requires knowing the RG of the pap), then realize that momentum will still be the same as it hits the pins.

The amount of momentum that is lost due to speed, is gained due to increased RPM.

There will be only one MPH and RPM combination that both meets the proper ratio (for a given axis tilt) and conserves the total momentum that was present at the release.